Example 4, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: The one-point compactification of $\mathbb{R}^2$ is homeomorphic with the unit sphere $S^2$

Question

Here is Theorem 29.1 in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be a (topological) space. Then $X$ is locally compact Hausdorff if and only if there exists a (topological) space $Y$ satisfying the following conditions:

(1) $X$ is a subspace of $Y$.

(2) The set $Y - X$ consists of a single point.

(3) $Y$ is a compact Hausdorff space.

If $Y$ and $Y^\prime$ are two spaces satisfying these conditions, then there is a homeomorphism of $Y$ with $Y^\prime$ that equals the identity map on $X$.

Immediately following the proof of Theorem 29.1, Munkres states

If $X$ itself should happen to be compact, then the space $Y$ of the preceding theorem is not very interesting, for it is obtained from $X$ by adjoining a single isolated point. However, if $X$ is not compact, then the point of $Y - X$ is a limit point of $X$, so that $\overline{X} = Y$.

Definition If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y - X$ equals a single point, then $Y$ is called the one-point compactification of $X$.

We have shown that $X$ has a one-point compactification $Y$ if and only if $X$ is a locally compact Hausdorff space that is not itself compact. We speak of $Y$ as "the" one-point compactification because $Y$ is uniquely determined up to a homeomorphism.

Here is my Math Stack Exchange post on the fact that the one-point compactification of the real line $\mathbb{R}$ is the unit circle $S^1$ (regarded as a subspace of $\mathbb{R}^2$).

Now my question is as follows:

Let $\mathbb{R}$ have the standard (or usual) topology, and let $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ have the product topology. Then how to show that the one-point compactification of $\mathbb{R}^2$ is (homeomorphic with) the unit sphere $S^2$ given by $$ S^2 = \left\{ (x, y, z) \in \mathbb{R}^3 \colon x^2 + y^2 + z^2 = 1 \right\}, $$ where $S^2$ is regarded as a subspace of $\mathbb{R}^3$? That is, can we find a point $P(a, b, c)$ on $S^2$ and a homeomorphism $f \colon \mathbb{R}^2 \longrightarrow S^2 \setminus \{ (a, b, c) \}$?

A supplementary question:

More generally, for each $n = 3, 4, 5, \ldots$, can we show that the one-point compactification of the euclidean space $\mathbb{R}^n$ is (homeomorphic with) the unit sphere $S^n \subset \mathbb{R}^{n+1}$ given by $$ S^n := \left\{ \, \left( x_1, \ldots, x_n, x_{n+1} \right) \in \mathbb{R}^{n+1} \, \colon \, x_1^2 + \cdots + x_n^2 + x_{n+1}^2 = 1 \, \right\}? $$ That is, can we find a point $P\left(a_1, \ldots, a_n, a_{n+1} \right) \in S^n$ and a homeomorphism $f \colon \mathbb{R}^n \longrightarrow S^n \setminus \left\{ \left( a_1, \ldots, a_n, a_{n+1} \right) \right\}$?

Answer 1

Let $X$ be a compact Hausdorff space and take a point $p\in X$ that is not isolated in $X$. Then $X\setminus\{p\}$ is locally compact, and is not compact (because it's not closed in $X$, because $p$ is not an isolated point of $X$). So $X$ itself is the one-point compactification of $X\setminus\{p\}$.

Applying this to your particular example, we have $X=S^n$, the unit sphere. If you remove one point from the sphere, you get a space homeomorphic to $\Bbb R^n$ by stereographic projection. So the one point compactification of the Euclidean space $\Bbb R^n$ is the sphere $S^n$ of the same dimension.

As far as showing the stereographic projection is a homeomorphism, it seems quite clear geometrically. One takes the unit sphere $S^n$ in the Euclidean space $\Bbb R^{n+1}$ and one projects from the "north pole" $Q=(1,0,\ldots,0)$ to the hyperplane containing the "equator". That maps a point $(x_0,x_1,...,x_n)\ne Q$ to a point $(0,X_1,...,X_n)$. It is clear geometrically that this is one-to-one and onto, and that if you move the point on the sphere slightly, the resulting projection to the hyperplane moves only slightly, and conversely, the reverse map is also continuous: if you move a point on the equator hyperplane slightly, the resulting intersection of the ray from that point to $Q$ with the sphere will also move slightly. That's all there is to it to prove it's a homeomorphism. No need for fancy formulas.

But if you really insist, the formulas are given in https://en.wikipedia.org/wiki/Stereographic_projection#Generalizations.

• For the forward map: $$X_i = \frac{x_i}{1 - x_0} \quad (i=1,...,n)$$

• For the inverse map, take $S^2=X_1^2+\cdots+X_n^2$ and define $$x_0=\frac{S^2-1}{S^2+1}\quad\text{and}\quad x_i=\frac{2 X_i}{S^2+1}\quad(i=1,...,n)$$